How do you factor 64x^3+8=0?

Mar 7, 2016

$8 \left(2 x + 1\right) \left(4 {x}^{2} - 2 x + 1\right) = 0$

Explanation:

This is a sum of cubes, since both $64 {x}^{3} = {\left(4 x\right)}^{3}$ and $8 = {\left(2\right)}^{3}$ are cubed terms.

Differences of cubes factor as follows:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

In $64 {x}^{3} + 8$, we see that $a = 4 x$ and $b = 2$, so

$64 {x}^{3} + 8 = 0$

$\left(4 x + 2\right) \left({\left(4 x\right)}^{2} - 4 x \left(2\right) + {\left(2\right)}^{2}\right) = 0$

$\left(4 x + 2\right) \left(16 {x}^{2} - 8 x + 4\right) = 0$

$8 \left(2 x + 1\right) \left(4 {x}^{2} - 2 x + 1\right) = 0$